Pythagoras Mod 5

Stage: 5 Challenge Level:

The problem was to prove that for every right angled triangle that
has sides with integer lengths

1. the area of the triangle is even and
2. the length of one of the sides is divisible by 5.

For both parts of the problem modulus arithmetic is useful. As a
hint for the first part, if $x$ is any integer then $x^2 \equiv 0,\
1$ (mod 4), that is it leaves a remainder of 0 or 1 when divided by
4. For the second you could use the fact that $x^2 \equiv -1,\ 0$
or 1 (mod 5).

The problem was solved by Tom from Dr.
Challoner's Grammar School, Amersham, Buckinghamshire.

The sides of any triangle are given by the triple $( x , y , z )$.
We only need to prove the result for primitive triples, that is in
which HCF$( x , y , z )=1$. If the area of such a triangle is even
and one side is a multiple of 5, then all other triangles with
triples which are multiples of primitive triples must have even
areas and a side which is a multiple of 5.

Say that $x^ 2 + y^ 2 = z^ 2$ and HCF$( x , y , z )=1.$ If $x$ and
$y$ are both even then $z$ is also even since an even number
squared is even and the sum of two even numbers is even. Hence
HCF$( x , y , z )> 1$ violating our supposition. If $x$ and $z$
are even then $y$ is even by similar logic (the difference of two
even numbers is even), also violating our supposition, and by
symmetry $z$ and $y$ cannot both be even. If $x$ and $y$ are both
odd then the sum of $x^2$ and $y^2$ is even, but of the form 2 (mod
4). Now all square numbers are of the form 1 (mod 4) or 0 (mod 4),
so $z^ 2$ is no longer a perfect square, violating the supposition
that $x ,\ y,\ z$ are integers.

Thus we can see that $x$ and $y$ are not both even or both odd,
hence one must be even, say without loss of generality $y$, and one
odd, say $x$. This implies that $z$ is odd as $x^2 + y^2 = 4( p^2 +
p )+1+4q^2\equiv 1$ (mod 4).

If $z$ is odd and equal to say $2r +1$, and $x = 2p +1$ while $y=
2q$, we see that: as $x^2 + y^2 = z^2$ it follows that $(2p+1)^2 +
(2q)^2 = (2r+1)^2$ so $q^2 = r(r+1) - p(p+1)$. But $p$ and $p +1$
are consecutive integers as are $r$ and $r+1$, hence one of each is
even, thus $p(p +1)$ is even as is $r(r+1$ and hence $q^2$ is even
which implies that $q$ is even. Thus $y$ is divisible by 4.

The area of a triangle is half base times height, that is $0.5 x
y$, and as $y$ is divisible by 4, then the area of a triangle with
a primitive Pythagorean triple is even, and by the logic earlier
all Pythagorean triples represent triangles with even areas.

In every case exactly one of the sides is a multiple of 5.
It should be noted that similar maths can be used to show that 3
divides one of $x ,\ y$ or $z$ in all primitive triples, and by the
logic at the beginning 5 divides one of $x ,\ y$ or $z$ for every
triple.

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